I have a $A$ Hermitian matrix that can be written as $A = B + iC$ where $i$ is an imaginary number. Also, $B$ and $C$ are $n$ dim real-value matrices. Now, there is a matrix,
$$
D = \begin{pmatrix} B & -C \\ C & B \end{pmatrix}
$$
Here, I would like to prove that $A$ and $D$ have the same eigenvalues in this case.
I have first attempted to prove that D is actually a symmetric matrix as below,
$$
\begin{align*}
A = A^* = \bar{A}^T \\
A = B + iC = (B – iC)^T, B = B^T, C = -C^T \\
D = \begin{pmatrix} B & -C \\ C & B \end{pmatrix} = \begin{pmatrix} B^T & C^T \\ -C^T & B^T \end{pmatrix} = D^T
\end{align*}
$$
However, I have no clue about writing the relation of eigenvalues with respect to matrices $A$ and $D$. Can anyone give me some suggestions for the next step? Thank you for your time!
Best Answer
In general, for any two real square matrices $B$ and $C$ of the same size, each eigenvalue $\lambda$ of $B+iC$ gives rise to a pair of eigenvalues $\lambda$ and $\overline{\lambda}$ of $D=\pmatrix{B&-C\\ C&B}$, because \begin{aligned} &\pmatrix{I&iI\\ 0&I}\pmatrix{B&-C\\ C&B}\pmatrix{I&iI\\ 0&I}^{-1}\\ &=\pmatrix{B+iC&-C+iB\\ C&B}\pmatrix{I&-iI\\ 0&I}\\ &=\pmatrix{B+iC&0\\ C&B-iC}\\ &=\pmatrix{B+iC&0\\ C&\overline{B+iC}}.\\ \end{aligned} When $B+iC$ is Hermitian, its eigenvalues are real. Hence $\lambda=\overline{\lambda}$ in the discussion above, meaning that the spectrum of $D$ is precisely two copies of the spectrum of $B+iC$.